BY similarity , there will be four cases in total. So the answer is 4 * 4C3 * 3! * 3 = 288 .

But the correct answer is 144. That means I am doubling up some number which I am unable to figure out.

Mar 30th 2009, 01:01 PM

steve23144

When you add the fourth ball to one of the three boxes you've added ball 1 to ball 2, for example, which is the same result as adding ball 2 to the combination when the box contained ball 1. Therefore all the combinations have been counted twice, so you need to halve the number.

Someone will no doubt correct me if I'm wrong.

Mar 30th 2009, 01:09 PM

Plato

Quote:

Originally Posted by champrock

The question is The number of ways in which 4 distinct balls can be put in 4boxes labeled a,b,c,d so that exactly one box remains empty:

ANSWER: Four times the number of surjections from a set of four to a set of three.
$\displaystyle Surj(4,3)=36$.

Mar 31st 2009, 01:28 AM

champrock

wat is a "Surjection" this seems a really scientific way to do it.

Searched on net but could not find its application as u have used in the above case.

Mar 31st 2009, 07:18 AM

Plato

Quote:

Originally Posted by champrock

what is a "Surjection" this seems a really scientific way to do it. Searched on net but could not find its application as u have used in the above case.

I find it hard to understand what you mean by “Searched on net but could not find its application”.
I found this nice page on one try.
A surjection is simply an onto function.
If you are serious about counting problems the you need to know about mappings.